Problem 24 - Calculus Early Transcendentals

Question

a. Determine whether the Mean Value Theorem applies to the following functions on the given interval \([a, b]\). b. If so, find the point(s) that are guaranteed to exist by the Mean Value Theorem. c. For those cases in which the Mean Value Theorem applies, make a sketch of the function and the line that passes through \((a, f(a))\) and\((b, f(b)) .\) Mark the points \(P\) at which the slope of the function equals the slope of the secant line. Then sketch the tangent line at \(P\). $$f(x)=x /(x+2) ;[-1,2]$$

Step-by-Step Solution

Answer

If so, what is the point guaranteed to exist by the MVT? Answer: Yes, the Mean Value Theorem can be applied to the function \(f(x) = \frac{x}{x + 2}\) on the interval \([-1, 2]\). The point guaranteed to exist by the MVT is \((0, 0)\).

Step 1: Verify if the conditions for MVT are met

To apply the Mean Value Theorem, the function must satisfy the following two conditions: 1. \(f(x)\) is continuous on \([a, b]\). 2. \(f(x)\) is differentiable on \((a, b)\). Our function \(f(x) = \frac{x}{x + 2}\) is a rational function. It will be continuous on its domain, so we need to check if there are any discontinuities in the domain of this function. The denominator of \(f(x)\) is \(x + 2\), which will have a discontinuity at \(x = -2\). Since \(-2\) is not in the interval \([-1, 2]\), \(f(x)\) is continuous on the given interval. Now, let's check if \(f(x)\) is differentiable on the interval \((-1, 2)\). To do that, we need to find the derivative of \(f(x)\).

Step 2: Find the derivative of f(x)

In this step, we will find the derivative of \(f(x)\). Using the quotient rule, we get: $$f'(x) = \frac{d}{dx} \frac{x}{x + 2} = \frac{(x + 2)(1) - x(1)}{(x + 2)^2} = \frac{2}{(x + 2)^2}.$$ Since the derivative \(f'(x)\) is defined for all \(x\) in the interval \((-1, 2)\), the function is differentiable on this interval. Thus, both conditions of MVT are met, and we can proceed to apply MVT to our function.

Step 3: Apply the Mean Value Theorem

According to the Mean Value Theorem, if \(f(x)\) is continuous on the interval \([a, b]\) and differentiable on \((a, b)\), then there exists at least one point \(c\) in \((a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). So let's find \(f'(c)\): $$f'(c) = \frac{2}{(c + 2)^2}.$$ Also, we need to find the values of \(f(a)\) and \(f(b)\): $$f(-1) = \frac{-1}{-1 + 2} = -1$$ $$f(2) = \frac{2}{2 + 2} = \frac{1}{2}$$ Now, we can compute the slope of the secant line: $$\frac{f(b) - f(a)}{b - a} = \frac{\frac{1}{2} - (-1)}{2 - (-1)} = \frac{\frac{3}{2}}{3} = \frac{1}{2}.$$

Step 4: Find the point(s) guaranteed by MVT

Now, by using the MVT result, we equate the derivative to the slope of the secant line: $$\frac{2}{(c + 2)^2} = \frac{1}{2}$$ To find the value(s) of \(c\), we can solve this equation: $$2 = \frac{1}{2}(c + 2)^2$$ $$(c + 2)^2 = 4$$ $$c + 2 = \pm 2$$ $$c = 0, -4$$ Since \(-4\) is not in the interval \((-1, 2)\), there is only one point guaranteed to exist by MVT which is \(c = 0\). To find the point on the curve, we plug \(c\) back into \(f(x)\): $$f(0) = \frac{0}{0 + 2} = 0$$ So, the point is \((0, 0)\).

Step 5: Sketch the function, secant line, and tangent line

Now that we know the parameters, draw the function \(f(x) = \frac{x}{x + 2}\) on the interval \([-1, 2]\), the secant line passing through \((-1, -1)\) and \((2, \frac{1}{2})\) with a slope of \(\frac{1}{2}\), and the tangent line at point \((0, 0)\) with a slope of \(\frac{1}{2}\). Label the intersection point between the function curve and the tangent line as point P. In summary, the Mean Value Theorem can be applied to the function \(f(x) = \frac{x}{x + 2}\) on the interval \([-1, 2]\). The points guaranteed to exist by the MVT are \((0, 0)\).

Key Concepts

Continuity

Continuity is a fundamental concept in calculus that ensures a function behaves smoothly. A function is considered continuous on a closed interval \([a, b]\) if there are no breaks, jumps, or holes in its graph within that interval. For the Mean Value Theorem (MVT) to apply, continuity is essential because it guarantees that the function can be appropriately analyzed over the desired range.
For the given function \(f(x) = \frac{x}{x + 2}\), we need to examine its behavior over the interval \([-1, 2]\). Since the function is a rational expression and the denominator \(x + 2\) can only be zero at \(x = -2\), which is outside our interval, the function is in fact continuous on \([-1, 2]\).
This continuity allows us to explore other important properties of the function in the context of MVT.

Differentiability

Differentiability describes whether a function has a derivative at each point within an open interval, like \((a, b)\). When a function is differentiable, its graph is smooth, and the slope can be calculated at every point.
The Mean Value Theorem requires differentiability on the open interval \((a, b)\). For our function \(f(x) = \frac{x}{x + 2}\), we calculate the derivative using the quotient rule, which yields \(f'(x) = \frac{2}{(x + 2)^2}\).
This derivative exists for all \(x\) in the interval \((-1, 2)\), confirming differentiability. The smoothness indicated by differentiability allows us to connect the rate of change in function values to the slope of the secant line, a crucial part of utilizing MVT.

Secant Line

The secant line is a straight line that connects two points on a function's graph, specifically at \((a, f(a))\) and \((b, f(b))\). It represents the average rate of change between those two points.
For our function, we calculate the slope of the secant line between \([-1, 2]\) as \(\frac{1}{2}\). This is done using the change in \(y\) over the change in \(x\), calculated by \(\frac{f(b) - f(a)}{b - a}\).
The secant line provides a visual and analytical reference for the Mean Value Theorem, helping us find the specific point where the function's instantaneous rate of change (slope of the tangent line) matches this average rate.

Tangent Line

A tangent line touches the curve of a function at exactly one point, providing the instantaneous rate of change (or slope) at that point. In the Mean Value Theorem, there must exist at least one point \(c\) in the open interval \((a, b)\) such that the slope of the tangent at \(c\) equals the slope of the secant line.
For our function, we found that \(f'(c) = \frac{1}{2}\) when \(c = 0\). This means at \((0, 0)\), the tangent line will have the same slope as our secant line, \(\frac{1}{2}\).
Drawing this tangent line helps visualize MVT, illustrating where the function doesn't just average the rate of change over the interval, but actually matches it at a specific point.